Question

In parallelogram $ABCD,$ point $M$ is on $\overline{AB}$ so that $\frac{AM}{AB} = \frac{17}{1000},$ and point $N$ is on $\overline{AD}$ so that $\frac{AN}{AD} = \frac{17}{2009}.$ Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}.$ Find $\frac{AC}{AP}.$

答案

Solution:

Place $A$ at the origin and let $\mathbf{b} = \overrightarrow{AB}$ and $\mathbf{d} = \overrightarrow{AD},$ so that $C = \mathbf{b} + \mathbf{d},$ $M = \frac{17}{1000}\mathbf{b},$ and $N = \frac{17}{2009}\mathbf{d}.$ Since $P$ lies on $\overline{AC},$ write $P = s\,(\mathbf{b} + \mathbf{d})$ where $s = \frac{AP}{AC};$ since $P$ also lies on line $MN,$ write $P = tM + (1 - t)N$ for some $t.$

Because $\mathbf{b}$ and $\mathbf{d}$ are independent, the coefficients must agree: $s = \frac{17t}{1000} \qquad \text{and} \qquad s = \frac{17(1-t)}{2009}.$ Thus $t = \frac{1000s}{17}$ and $1 - t = \frac{2009s}{17};$ adding gives $1 = \frac{3009s}{17}.$

Therefore $\frac{AC}{AP} = \frac{1}{s} = \frac{3009}{17} = 177.$

2009 AIME I Problem 4

Solution:

Problem 4 in Other Years